Question

Sol.

(A) R = 2 Vsin$\frac{\theta}{2}$

= 2 × 10 × sin60

= 20 × $\frac{\sqrt{3}}{2}$ = 10$\sqrt{3}$.

(B) a_t = 0.

a_c = a = $\frac{v^{2}}{R} = \frac{100}{10}$ = 10 m/s²

(C) Avg. acceleration = $\frac{changeininvelocity}{time}$

θ = $\frac{Arc}{Raduis}$ Now; time = $\frac{Arc}{speed} = \frac{20\pi}{3 \times 10} = \frac{2\pi}{3}$sec.

$\frac{2\pi}{3} \times 10 = Arc$

$\frac{20\pi}{3} = Arc$

Change in velocity = $(5\widehat{j} - 5\sqrt{3}\widehat{i}) - 10\widehat{j}$

= – $(5\sqrt{3}\widehat{i} + 5\widehat{j})$

= $- \frac{(5\sqrt{3}\widehat{i} + 5\widehat{j})}{\sqrt{(5\sqrt{3})^{2} + (5)^{2})}}$

Now = $- \frac{(5\sqrt{3}\widehat{i} + 5\widehat{j}) \times 3}{10 \times 2\pi}$

(D) No it’s not same bcz Inst. acc. = $\frac{dv}{dt}$

Avg acc. = $\frac{\Delta v}{\Delta t}$